Find the gradient of $f(x, y, z) = yx + \sin(z)$ at $(1, 2, \pi)$. $\nabla f(1, 2, \pi) = ($
Explanation: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 3D scalar field, this looks like $\nabla f = (f_x, f_y, f_z)$. Let's find $f_x$, $f_y$, and $f_z$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ yx + \sin(z) \right] \\ \\ &= y \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ yx + \sin(z) \right] \\ \\ &= x \\ \\ f_z &= \dfrac{\partial}{\partial z} \left[ yx + \sin(z) \right] \\ \\ &= \cos(z) \end{aligned}$ Now we can evaluate the partial derivatives we found at the point $(1, 2, \pi)$. $\begin{aligned} f_x(1, 2, \pi) &= y = 2 \\ \\ f_y(1, 2, \pi) &= x = 1 \\ \\ f_z(1, 2, \pi) &= \cos(z) = -1 \end{aligned}$ The gradient of $f$ at $(1, 2, \pi)$ is $\nabla f = (2, 1, -1)$.